how about something like this ....
n!/(2n)!
= n(n-1)(n-2)(n-3) ... (2)(1)/ [(2n)(2n-1)(2n-2)(2n-3) ..(2)(1) ]
= n(n-1)(n-2)(n-3) ... (2)(1)/ 2n(2n-1)(2)(n-1)(2n-3)(2)(n-2)(2n-5)(2)(n-3) ... (2)(1) ]
notice that all of the factors of the numerator are found in the denominator, so ...
= 1/ [ 2^n(2n-1)(2n-3)(2n-5) ... (2)(1) ]
clearly as n ---> ∞, the denominator --- ∞
1/(very large) ---> 0
and the limit n!/(2n)! = 0
Prove that the limit as n approaches infinity of ((n!)^2)/(2n)! is 0.
I'm thinking that you have to use the squeeze theorem, but I'm not quite sure how. help please?
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